An automobile with a mass of 961.000 kg has 3.27 m between the front and rear axles. Its center of gravity is located 0.615 m behind the front axle. With the automobile on level ground, determine the magnitude of the force from the ground on (a) each front wheel (assuming equal forces on the front wheels) and (b) each rear wheel (assuming equal forces on the rear wheels).
Question
Answer:
a) 3830 N
b) 887 N
Explanation:
WOW, you know the mass to within ± ½gram and position of CG to ± 0.5 mm! that’s some impressive measuring!!!
Let FR be the total rear wheels force
Let FF be the total front wheels force
Sum moments about the front wheels to zero
961(9.81)[0.615] – FR[3.27] = 0
FR = 1,773.045
so each rear tire supports
Fr = 1,773.045/2 = 886.5225 N
Fr ≈ 887 N when rounded to three significant digits
Sum moments about the rear wheels to zero
961(9.81)[3.27 – 0.615] – FF[3.27] = 0
FF = 7,654.365
so each front tire supports
Ff = 7,654.365/2 = 3,827.1825
Ff = 3830 N (to 3 s.d.)
verify by summing vertical forces to zero
7,654.365 + 1,773.045 – 961(9.81) ?=? 0
0 = 0 checks out